Description：

You have a bag of size $n$. Also you have m boxes. The size of $i-th$ box is $a_{i}$, where each ai is an integer non-negative power of two.

You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.

For example, if $n=10$ and $a=[1,1,32]$ then you have to divide the box of size 32 into two parts of size $16$, and then divide the box of size 16. So you can fill the bag with boxes of size $1, 1$ and $8$.

Calculate the minimum number of divisions required to fill the bag of size n.

Input

The first line contains one integer $t (1≤t≤1000)$ — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ $(1≤n≤10^{18},1≤m≤10^5)$ — the size of bag and the number of boxes, respectively.

The second line of each test case contains m integers $a_{1},a_{2},…,a_{m} (1≤a_{i}≤10^9)$ — the sizes of boxes. It is guaranteed that each ai is a power of two.

It is also guaranteed that sum of all m over all test cases does not exceed $10^5$.

Output

For each test case print one integer — the minimum number of divisions required to fill the bag of size $n$ (or −1, if it is impossible).

Example

input

3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8

output

2
-1
0

题意：

给你一个序列，问你能不能组合出 $n$，你可以把其中一个数变成两个大小相等的部分，问你最少的操作次数。
一种情况不能组成那就是全部加起来还是小于 $n$ ，然后我们就开始贪心，从大到小如果这个数比背包质量小那就直接放进去，如果比背包质量大并且剩下的质量填不满背包我们就操作一次，否则我们舍弃这个物品。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

ll n;
ll m;
const int N = 1e5 + 10;
ll a[N], sum, ans;
int main()
{
    int t;
    sd(t);
    while (t--)
    {
        sldd(n, m);
        sum = 0;
        rep(i, 1, m)
        {
            sld(a[i]);
            sum += a[i];
        }
        if (sum < n)
        {
            puts("-1");
            continue;
        }
        sort(a + 1, a + m + 1);
        ans = 0;
        while (m)
        {
            if (a[m] <= n)
            {
                n -= a[m];
                sum -= a[m--];
            }
            else if (sum - a[m] < n)
            {
                a[m] /= 2;
                a[m + 1] = a[m];
                m++;
                ans++;
            }
            else
                sum -= a[m--];
            //cout << m << " " << a[m] << endl;
        }
        pld(ans);
    }
    return 0;
}