Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int n,T,b[200003],dp[200003],a[200003],dat[500003];
vector<int>v[200003];
void upd(int l,int r,int x){
	while(l<=r){
		dat[l]=min(dat[l],x);
		dat[r]=min(dat[r],x);
		l=(l+1)/2;
		r=(r-1)/2;
	}
}
int fnd(int x){
	int ret=1234567;
	while(x){
		ret=min(ret,dat[x]);
		x/=2;
	}
	return ret;
} 
int main(){
	memset(dat,31,sizeof(dat));
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",b+i);
	scanf("%d",&T);
	for(int i=1;i<=n;i++)a[i]=a[i-1]+1-b[i];
	while(T--){
		int l,r;scanf("%d%d",&l,&r);
		v[r].push_back(l);
	}
	for(int i=1;i<=n;i++)
	{
		upd(i+262143,i+262143,dp[i-1]-a[i-1]);
		dp[i]=dp[i-1]+b[i];
		for(int j=0;j<v[i].size();j++)
			upd(v[i][j]+262144,i+262144,fnd(v[i][j]+262143));
		dp[i]=min(dp[i],a[i]+fnd(i+262144));
	}
	cout<<dp[n];
}