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>
#define PB push_back
using namespace std;
typedef long long LL;
const int MAXN=2e5 + 10, SEGSZ = 262144, oo= 1<<29;

int zkw[SEGSZ*2+10],N, w[MAXN], pre[MAXN], f[MAXN], Q;
vector<int> G[MAXN];
template<class T> void upmin(T &a ,T b){if( a>b) a=b;}
void Modify(int l, int r, int x){	for (l+=SEGSZ, r+=SEGSZ; l<=r; l=(l+1)>>1, r=(r-1)>>1) upmin(zkw[l],x), upmin(zkw[r],x);}
int Query(int p){
	int res =oo ;
	for (p+=SEGSZ; p ; p>>=1 ) upmin(res, zkw[p]);
	return res;
}

int main(){
	scanf("%d", &N);
	for (int i=0; i< SEGSZ<<1; i++) zkw[i] =oo;
	for (int i=1; i<=N; i++){
		scanf("%d", &w[i]);
		pre[i] = pre[i-1] +1-w[i];
	}
	scanf("%d", &Q);
	for (int b,e; Q--;){
		scanf("%d%d", &b, &e);
		G[e].PB(b);
	}
	for (int i=1; i<=N; i++){
		Modify(i-1,i-1,f[i-1]-pre[i-1]);
		for (int j=0,v; j<G[i].size()?v=G[i][j],1:0; j++)	Modify(v, i, Query(v-1));
		f[i] = min(f[i-1]+w[i] , Query(i)+pre[i]);
	}
	printf("%d\n", f[N]);
	return 0;
}