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>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll mod=(1e9)+7;
const ll iv=(5e8)+4;
int n,q,a[3010],x,y;
ll dp[3010][3010],tmp=1,ans;
int main() {
	read(n); read(q);
	for (int i=1;i<=n;i++) read(a[i]);
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
		if (a[i]>a[j]) dp[i][j]=1;
	while (q--) {
		read(x); read(y); tmp=tmp*2%mod;
		for (int i=1;i<=n;i++)
			if (i!=x&&i!=y) {
				dp[i][x]=dp[i][y]=(dp[i][x]+dp[i][y])*iv%mod;
				dp[x][i]=dp[y][i]=(dp[x][i]+dp[y][i])*iv%mod;
			}
		dp[x][y]=dp[y][x]=(dp[x][y]+dp[y][x])*iv%mod;
	}
	for (int i=1;i<=n;i++)
	for (int j=i+1;j<=n;j++)
		ans+=dp[i][j],ans%=mod;
	printf("%lld\n",ans*tmp%mod);
	return 0;
}