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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007,inv2=500000004;
inline void add(int &x,int y){(x+=y)>=mod?x-=mod:0;}
inline int pl(int x,int y){return (x+=y)>=mod?x-mod:x;}
int n,q,a[3030],x[3030],y[3030],dp[3030][3030],tmp[3030][3030],ans;
int main()
{
	scanf("%d %d",&n,&q);
	for(int i=1;i<=n;++i)scanf("%d",&a[i]);
	for(int i=1;i<=q;++i)scanf("%d %d",&x[i],&y[i]);
	for(int i=1;i<=n;++i)for(int j=1;j<i;++j)dp[i][j]=1;
	for(int i=q;i;--i)
	{
		for(int j=1;j<=n;++j)
		{
			tmp[x[i]][j]=1ll*dp[x[i]][j]*inv2%mod;
			tmp[y[i]][j]=1ll*dp[y[i]][j]*inv2%mod;
			tmp[j][x[i]]=1ll*dp[j][x[i]]*inv2%mod;
			tmp[j][y[i]]=1ll*dp[j][y[i]]*inv2%mod;
		}
		for(int j=1;j<=n;++j)
		{
			dp[y[i]][j]=dp[x[i]][j]=pl(tmp[x[i]][j],tmp[y[i]][j]);
			dp[j][x[i]]=dp[j][y[i]]=pl(tmp[j][x[i]],tmp[j][y[i]]);
		}
		dp[x[i]][y[i]]=dp[y[i]][x[i]]=pl(tmp[x[i]][y[i]],tmp[y[i]][x[i]]);
	}
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			if(a[i]<a[j])add(ans,dp[i][j]);
	for(int i=1;i<=q;++i)add(ans,ans);
	printf("%d",ans);
	return 0;
}