You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types: 

  * increase x by 1; 
  * decrease x by 1. 



You are given m queries of the following format: 

  * query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order? 

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Then the description of t testcases follows.

The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.

The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.

Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5. 

Output

For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.

Example

Input


2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4


Output


1
2
4
4
3
3
4
2
3
2
1
2
2
2

Note

The instructions that remain for each query of the first testcase are: 

  1. empty program — x was only equal to 0; 
  2. "-" — x had values 0 and -1; 
  3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them; 
  4. "+--+--+" — the distinct values are 1, 0, -1, -2. 
#include <bits/stdc++.h>
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
using namespace std;

typedef long long ll;
typedef unsigned long long llu;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<int, pii> piii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;
typedef pair<string, int> psi;
typedef pair<char, int> pci;
typedef pair<int, char> pic;
const int MOD = 1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;

ll fac[1] = {1}, inv[1] = {1};
ll gcd(ll a,ll b){return abs(b)?gcd(abs(b),abs(a)%abs(b)):abs(a);}
ll mp(ll a,ll b){ll ret=1;while(b){if(b&1)ret=ret*a%MOD;a=a*a%MOD;b>>=1;}return ret;}
ll cmb(ll r, ll c) {return (c>r||c<0) ? 0:fac[r] * inv[c] % MOD * inv[r - c] % MOD;}

char s[200005];
int ps[200005];
int lmi[200005], lma[200005];
int rmi[200005], rma[200005];

int main() {
    int tc;
    scanf("%d", &tc);

    while (tc--) {
        int n, m;
        scanf("%d %d", &n, &m);
        scanf("%s", s + 1);

        int sum = 0;
        for (int i = 1; i <= n; i++) {
            if (s[i] == '-') s[i] = -1;
            else s[i] = 1;
            lmi[i] = 1e9, lma[i] = -1e9;
            rmi[i] = 1e9, rma[i] = -1e9;
        }

        for (int i = 1; i <= n; i++) {
            sum += s[i];
            ps[i] = sum;
            lmi[i] = min(ps[i], lmi[i - 1]);
            lma[i] = max(ps[i], lma[i - 1]);
        }

        rmi[n + 1] = rma[n + 1] = sum;
        for (int i = n; i >= 1; i--) {
            rmi[i] = min(ps[i], rmi[i + 1]);
            rma[i] = max(ps[i], rma[i + 1]);
        }

        while (m--) {
            int l, r;
            scanf("%d %d", &l, &r);

            int mav = -1e9, miv = 1e9;
            mav = max(mav, lma[l - 1]);
            miv = min(miv, lmi[l - 1]);

            // printf("%d // %d\n", mav, miv);
            int t = ps[r] - ps[l - 1];
            mav = max(mav, rma[r + 1] - t);
            miv = min(miv, rmi[r + 1] - t);

            // printf("%d // %d\n", mav, miv);

            printf("%d\n", mav - miv + 1);
        }
    }
}

// author: rdd6584