#include <bits/stdc++.h>
#define int long long
#define INF 1e+9
using namespace std;

int seg = 1,segMin[600000],segAdd[600000];

void add(int a,int b,int x,int k = 0,int l = 0,int r = seg){
	if(b <= l || r <= a) return;
	if(a <= l && r <= b){
		segAdd[k] += x;
		return;
	}
	add(a,b,x,k * 2 + 1,l,(l + r) / 2);
	add(a,b,x,k * 2 + 2,(l + r) / 2,r);
	segMin[k] = min(segAdd[k * 2 + 1] + segMin[k * 2 + 1],segAdd[k * 2 + 2] + segMin[k * 2 + 2]);
}

int get(int a,int b,int k = 0,int l = 0,int r = seg){
	if(b <= l || r <= a) return INF;
	if(a <= l && r <= b) return segMin[k] + segAdd[k];
	return min(get(a,b,k * 2 + 1,l,(l + r) / 2),get(a,b,k * 2 + 2,(l + r) / 2,r)) + segAdd[k];
}

signed main(){
	int n,q,b[200010];
	vector<int> vec[200010];
	cin >> n;
	while(seg < n + 1) seg *= 2;
	add(0,seg,INF);
	add(0,1,-INF);
	for(int i = 0;i < n;i++) cin >> b[i];
	cin >> q;
	for(int i = 0;i < q;i++){
		int l,r;
		cin >> l >> r;
		vec[l - 1].push_back(r);
	}
	for(int i = 0;i < n;i++){
		for(int v : vec[i]) add(v,v + 1,get(0,v + 1) - get(v,v + 1));
		if(b[i]) add(0,i + 1,1);
		else add(i + 1,seg,1);
	}
	cout << get(0,seg) << endl;
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define int long long
#define INF 1e+9
using namespace std;

int seg = 1,segMin[600000],segAdd[600000];

void add(int a,int b,int x,int k = 0,int l = 0,int r = seg){
	if(b <= l || r <= a) return;
	if(a <= l && r <= b){
		segAdd[k] += x;
		return;
	}
	add(a,b,x,k * 2 + 1,l,(l + r) / 2);
	add(a,b,x,k * 2 + 2,(l + r) / 2,r);
	segMin[k] = min(segAdd[k * 2 + 1] + segMin[k * 2 + 1],segAdd[k * 2 + 2] + segMin[k * 2 + 2]);
}

int get(int a,int b,int k = 0,int l = 0,int r = seg){
	if(b <= l || r <= a) return INF;
	if(a <= l && r <= b) return segMin[k] + segAdd[k];
	return min(get(a,b,k * 2 + 1,l,(l + r) / 2),get(a,b,k * 2 + 2,(l + r) / 2,r)) + segAdd[k];
}

signed main(){
	int n,q,b[200010];
	vector<int> vec[200010];
	cin >> n;
	while(seg < n + 1) seg *= 2;
	add(0,seg,INF);
	add(0,1,-INF);
	for(int i = 0;i < n;i++) cin >> b[i];
	cin >> q;
	for(int i = 0;i < q;i++){
		int l,r;
		cin >> l >> r;
		vec[l - 1].push_back(r);
	}
	for(int i = 0;i < n;i++){
		for(int v : vec[i]) add(v,v + 1,get(0,v + 1) - get(v,v + 1));
		if(b[i]) add(0,i + 1,1);
		else add(i + 1,seg,1);
	}
	cout << get(0,seg) << endl;
	return 0;
}
```