Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N=3e3+5;
const int mod=1e9+7;
const int inv2=500000004;
int n,x,y,q,ans,tmp=1,a[N],f[N][N];

template <typename _Tp> inline void IN(_Tp&x) {
	char ch;bool flag=0;x=0;
	while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	if(flag) x=-x;
}

int main() {
	IN(n),IN(q);
	for(int i=1;i<=n;++i) IN(a[i]);
	for(int i=1;i<=n;++i) for(int j=1;j<=n;++j) f[i][j]=(bool)(a[i]>a[j]);
	for(int t=1;t<=q;++t) {
		IN(x),IN(y),
		f[x][y]=f[y][x]=1ll*(f[x][y]+f[y][x])*inv2%mod;
		for(int i=1;i<=n;++i) if(i!=x&&i!=y)
			f[i][x]=f[i][y]=1ll*(f[i][x]+f[i][y])*inv2%mod,
			f[x][i]=f[y][i]=1ll*(f[x][i]+f[y][i])*inv2%mod;
		tmp=2ll*tmp%mod;
	}
	for(int i=1;i<=n;++i) for(int j=i+1;j<=n;++j) ans=(ans+f[i][j])%mod;
	printf("%d\n",1ll*ans*tmp%mod);
	return 0;
}