#include<bits/stdc++.h>
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
#define ED(x) ((int)x.size()-1)
#define _for(i,a,b) for(int i=(a);i<=(b);++i)
#define _rep(i,a,b) for(int i=(a);i>=(b);--i)
#define changxv ios::sync_with_stdio(0);cin.tie(0);

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;

const int maxn=3003;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;

int f[maxn][maxn];
int s[maxn],r[maxn];

int main(){changxv
	int n,m;cin>>n>>m;
	_for(i,1,n){
		char c;cin>>c;
		s[i]=s[i-1]+c-'0';
	}
	s[n+1]=s[n];
//	_for(i,1,n)cerr<<s[i]<<' ';
//	cerr<<'\n';
	_for(i,1,n+1)r[i]=i;
	_for(i,1,m){
		int x,y;cin>>x>>y;
		r[x]=max(r[x],y);
	}
	_for(i,1,n+1)r[i]=max(r[i],r[i-1]);
	
	f[1][s[r[1]]]=1;
	_for(i,1,n){
		_for(j,0,n)if(f[i][j]){
			int L=r[i]+1,R=r[i+1];
			if(j)(f[i+1][j-1+s[R]-s[L-1]]+=f[i][j])%=mod;
			if(r[i]-i+1-j)(f[i+1][j+s[R]-s[L-1]]+=f[i][j])%=mod;
		}
	}
	cout<<f[n+1][0]<<'\n';
return 0; }
### Prompt

Create a solution in cpp for the following problem:

There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

### Response

```cpp
#include<bits/stdc++.h>
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
#define ED(x) ((int)x.size()-1)
#define _for(i,a,b) for(int i=(a);i<=(b);++i)
#define _rep(i,a,b) for(int i=(a);i>=(b);--i)
#define changxv ios::sync_with_stdio(0);cin.tie(0);

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;

const int maxn=3003;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;

int f[maxn][maxn];
int s[maxn],r[maxn];

int main(){changxv
	int n,m;cin>>n>>m;
	_for(i,1,n){
		char c;cin>>c;
		s[i]=s[i-1]+c-'0';
	}
	s[n+1]=s[n];
//	_for(i,1,n)cerr<<s[i]<<' ';
//	cerr<<'\n';
	_for(i,1,n+1)r[i]=i;
	_for(i,1,m){
		int x,y;cin>>x>>y;
		r[x]=max(r[x],y);
	}
	_for(i,1,n+1)r[i]=max(r[i],r[i-1]);
	
	f[1][s[r[1]]]=1;
	_for(i,1,n){
		_for(j,0,n)if(f[i][j]){
			int L=r[i]+1,R=r[i+1];
			if(j)(f[i+1][j-1+s[R]-s[L-1]]+=f[i][j])%=mod;
			if(r[i]-i+1-j)(f[i+1][j+s[R]-s[L-1]]+=f[i][j])%=mod;
		}
	}
	cout<<f[n+1][0]<<'\n';
return 0; }
```