#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
typedef long long ll;
#define mod 1000000007
#define N 3050
inline void upd(ll &x,ll y) {x=(x+y)%mod;}
ll f[N][N];
int s[N],n,m,rp[N];
char w[N];
int main() {
	scanf("%d%d%s",&n,&m,w+1);
	int i,j;
	for(i=1;i<=n;i++) s[i]=s[i-1]+w[i]-'0';
	s[n+1]=s[n];
	for(i=1;i<=n+1;i++) rp[i]=i;
	for(i=1;i<=m;i++) {
		int l,r;
		scanf("%d%d",&l,&r);
		rp[l]=max(rp[l],r);
	}
	for(i=1;i<=n;i++) rp[i]=max(rp[i],rp[i-1]);
	f[1][s[rp[1]]]=1;
	for(i=1;i<=n;i++) {
		for(j=0;j<=n;j++) if(f[i][j]) {
			int num=s[rp[i+1]]-s[rp[i]];
			if(j) upd(f[i+1][j-1+num],f[i][j]);
			if(j!=rp[i]-i+1) upd(f[i+1][j+num],f[i][j]);
		}
	}
	printf("%lld\n",f[n+1][0]);
}

### Prompt

Please create a solution in CPP to 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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
typedef long long ll;
#define mod 1000000007
#define N 3050
inline void upd(ll &x,ll y) {x=(x+y)%mod;}
ll f[N][N];
int s[N],n,m,rp[N];
char w[N];
int main() {
	scanf("%d%d%s",&n,&m,w+1);
	int i,j;
	for(i=1;i<=n;i++) s[i]=s[i-1]+w[i]-'0';
	s[n+1]=s[n];
	for(i=1;i<=n+1;i++) rp[i]=i;
	for(i=1;i<=m;i++) {
		int l,r;
		scanf("%d%d",&l,&r);
		rp[l]=max(rp[l],r);
	}
	for(i=1;i<=n;i++) rp[i]=max(rp[i],rp[i-1]);
	f[1][s[rp[1]]]=1;
	for(i=1;i<=n;i++) {
		for(j=0;j<=n;j++) if(f[i][j]) {
			int num=s[rp[i+1]]-s[rp[i]];
			if(j) upd(f[i+1][j-1+num],f[i][j]);
			if(j!=rp[i]-i+1) upd(f[i+1][j+num],f[i][j]);
		}
	}
	printf("%lld\n",f[n+1][0]);
}

```