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. 
/*
17/10/2017
*/
#include <bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;

struct segtree {
    ll size;
    vector<ll> sums;
    void init(ll n) {
        size = 1;
        while (size < n) {
            size *= 2;
        }
        sums.assign(2 * size, -1e9);
    }
    void build(vector<ll> &a, ll x, ll lx, ll rx) {
        if (rx - lx == 1) {
            if (lx < a.size()) {
                sums[x] = a[lx];
            }
            return;
        }
        ll m = (lx + rx) / 2;
        build(a, 2 * x + 1, lx, m);
        build(a, 2 * x + 2, m, rx);
        sums[x] = max(sums[2 * x + 1], sums[2 * x + 2]);
    }
    void build(vector<ll> &a) {
        build(a, 0, 0, size);
    }
    ll sum(ll l, ll r, ll x, ll lx, ll rx) {
        if (lx >= r || rx <= l) {
            return -1e9;
        }
        if (lx >= l && rx <= r) {
            return sums[x];
        }
        ll m = (lx + rx) / 2;
        ll s1 = sum(l, r, 2 * x + 1, lx, m);
        ll s2 = sum(l, r, 2 * x + 2, m, rx);
        return max(s1, s2);
    }
    ll sum(ll l, ll r) {
        return sum(l, r, 0, 0, size);
    }
    void set(ll i, ll v, ll x, ll lx, ll rx) {
        if (rx - lx == 1) {
            sums[x] = v;
            return;
        }
        ll m = (lx + rx) / 2;
        if (i < m) {
            set(i, v, 2 * x + 1, lx, m);
        } else {
            set(i, v, 2 * x + 2, m, rx);
        }
        sums[x] = max(sums[2 * x + 1], sums[2 * x + 2]);
    }
    void set(ll i, ll v) {
        set(i, v, 0, 0, size);
    }
};


int main()
{
ll t;
cin>>t;
while(t--)
{
ll n,m;
cin>>n>>m;
string s;
cin>>s;
ll ans=0;
vector<ll> pref;
vector<pair<ll,ll> > ranges;
//pref.push_back(0);
ll sm=0;
segtree mxs,mns;
mxs.init(n);
mns.init(n);
for(ll i=0;i<n;i++)
{
sm+=(s[i]=='+');
sm-=(s[i]=='-');
mxs.set(i,sm);
mns.set(i,-sm);
pref.push_back(sm);
}
ll l=0,r=0,c=0;
for(ll i=0;i<n;i++)
{
c+=(s[i]=='+');
c-=(s[i]=='-');
if(c>r)
{
r++;
}
if(c<l)
{
l--;
}
ranges.push_back({l,r});
}
while(m--)
{
ll x,y;
cin>>x>>y;
ll ans=0;
x--,y--;
l=0,r=0;
ll mn=-mns.sum(y+1,n);
ll mx=mxs.sum(y+1,n);
ll p=0,f=0;
p=pref[y];
if(x-1>=0)
{
f=pref[x-1];
}
if(mn!=1e9)
{
l=mn-p+f;
r=mx-p+f;
}
l=min(l,0ll);
//cout<<l<<" "<<r<<endl;
ll ix=0,iy=0;
if(x-1>=0)
{
ix=ranges[x-1].first,iy=ranges[x-1].second;
}
cout<<iy-ix+1+r-l+1-max(0ll,-max(ix,l)+min(iy,r)+1)<<endl;
}
}
}
