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:
/*

*/ 
//计数->期望/概率->独立 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int inv_2=5e8+4;
const int N=3005;
int dp[N][N],a[N],x,y,ans,n,q;
void up(int &x,int y){x=x-mod+y;x=(x<0)?x+mod:x;}
int ksm(int x,int y)
{
	int res=1;
	while (y)
	{
		if (y&1) res=(ll)res*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}
	return res;
}
void SWAP(int x,int y)
{
	for (int i=1;i<=n;i++)
	{
		if (i==x||i==y)
		{
			dp[x][y]=dp[y][x]=(ll)(dp[x][y]+dp[y][x])*inv_2%mod;
		}else{
			dp[i][x]=dp[i][y]=(ll)(dp[i][x]+dp[i][y])*inv_2%mod;
			dp[x][i]=dp[y][i]=(ll)(dp[x][i]+dp[y][i])*inv_2%mod;
		}
	}
}
int main()
{
	scanf("%d%d",&n,&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];
	for (int i=1;i<=q;i++)
	{
		scanf("%d%d",&x,&y);
		if (x>y) swap(x,y);
		SWAP(x,y);
	}	
	int t=ksm(2,q);
	for (int i=1;i<=n;i++)
	  for (int j=i+1;j<=n;j++) up(ans,(ll)dp[i][j]*t%mod);
	printf("%d\n",ans);
	return 0;
}