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 Mod 1000000007
#define inv 500000004
#define N 5005
using namespace std;
int n,x,y,Q,a[N],dp[N][N];
inline int ksm(int x,int y){
	int ans1=1;while (y){
		if (y&1) ans1=1ll*ans1*x%Mod;
		y>>=1;x=1ll*x*x%Mod;
	}return ans1;
} 
int main(){
	scanf("%d%d",&n,&Q);int tmp=ksm(2,Q); 
	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++) dp[i][j]=a[i]<a[j];//a[i]<a[j]的方案数 
	while (Q--){
		scanf("%d%d",&x,&y);
		if (x>y) swap(x,y);
		for (int i=1;i<=n;i++){
			if (i==x||i==y) continue;
			//dp[x][i]=dp[y][i]=(dp[x][i]+dp[y][i])%Mod;
			//dp[i][x]=dp[i][y]=(dp[i][x]+dp[i][y])%Mod;
			dp[x][i]=dp[y][i]=1ll*(dp[x][i]+dp[y][i])*inv%Mod;
			dp[i][x]=dp[i][y]=1ll*(dp[i][x]+dp[i][y])*inv%Mod;
		}
		//dp[x][y]=dp[y][x]=(dp[x][y]+dp[y][x])%Mod;
		dp[x][y]=dp[y][x]=1ll*(dp[x][y]+dp[y][x])*inv%Mod;
	}
	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=i;j++) ans=(ans+1ll*tmp*dp[i][j]%Mod)%Mod;
	printf("%d\n",ans); 
	return 0;
}