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<bits/stdc++.h>
#define tr(xx,yy) (xx+=yy)>=M?xx-=M:0
const int N=3003,M=1000000007,I2=M+1>>1;
using namespace std;
int n,m,a[N],f[N][N],l,r,
g0[N],g1[N],h0[N],h1[N],ans;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	scanf("%d",a+i);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	if(a[j]<a[i])f[j][i]=1;
	for(int al=0;al<m;al++){
		scanf("%d%d",&l,&r);
		for(int i=1;i<=n;i++)
		if(i!=r)
		f[l][i]=1ll*f[l][i]*I2%M,
		f[i][l]=1ll*f[i][l]*I2%M;
		for(int i=1;i<=n;i++)
		f[r][i]=1ll*f[r][i]*I2%M,
		f[i][r]=1ll*f[i][r]*I2%M;
		for(int i=1;i<=n;i++)
		g0[i]=f[l][i],g1[i]=f[i][l],
		h0[i]=f[r][i],h1[i]=f[i][r];
		for(int i=1;i<=n;i++){
			if(i!=r)tr(f[r][i],g0[i]),tr(f[i][r],g1[i]);
			else tr(f[r][l],g0[i]),tr(f[l][r],g1[i]);
			if(i!=l)tr(f[l][i],h0[i]),tr(f[i][l],h1[i]);
		}
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<i;j++)tr(ans,f[i][j]);
	for(int i=0;i<m;i++)tr(ans,ans);
	printf("%d",ans);
}