#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 2333333
#define MIN(_A, _B) (((_A) < (_B)) ? (_A) : (_B))
 
int n, Q;
int B[233333];
 
struct Data{
	int l, r;
	bool operator <(const Data a)const{
		if(l == a.l) return r > a.r;
		return l < a.l;
	}
}d[233333];
 
 
//Seg_Tree
int ql, qr;
int t[999999];
int query(int now, int l, int r){
	if(l >= ql && r <= qr) return t[now];
	int mid = l + r >> 1, qa = INF, qb = INF;
	if(ql <= mid) qa = query(now << 1, l, mid);
	if(qr > mid) qb = query(now << 1 | 1, mid + 1, r);
	return MIN(qa, qb);
}
void modifm(int now, int l, int r, int q, int v){
	if(l == r) {t[now] = MIN(t[now], v); return;} 
	int mid = l + r >> 1;
	if(q <= mid) modifm(now << 1, l, mid, q, v);
	else modifm(now << 1 | 1, mid + 1, r, q, v);
	t[now] = MIN(t[now << 1], t[now << 1 | 1]);
}
 
int ans = 0;
void input(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		scanf("%d", &B[i]), (B[i] == 0) && (ans++);
	scanf("%d", &Q);
	for(int i = 1; i <= Q; i++)
		scanf("%d %d", &d[i].l, &d[i].r);
	std::sort(d + 1, d + Q + 1);
	memset(t, 63, sizeof(t));
}
 
void work(){
	int now = 1;
	for(int i = 1, tmp = 0; i <= n; i++){
		while(now <= Q && d[now].l == i){
			ql = i - 1, qr = d[now].r;
			tmp = query(1, 1, n);
			if(ql == 0) tmp = MIN(tmp, 0);
			modifm(1, 1, n, qr, tmp);
			now++;
		}
		ql = qr = i - 1;
		if(i == 1) tmp = 0;
		else tmp = query(1, 1, n);
		(B[i] == 0) ? (tmp--) : (tmp++);
		modifm(1, 1, n, i, tmp);
	}
	ql = qr = n;
	printf("%d\n", ans + query(1, 1, n));
}
 
int main(){
	input();
	work();
	return 0;
}
### Prompt

Please formulate 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 <cstdio>
#include <cstring>
#include <algorithm>
#define INF 2333333
#define MIN(_A, _B) (((_A) < (_B)) ? (_A) : (_B))
 
int n, Q;
int B[233333];
 
struct Data{
	int l, r;
	bool operator <(const Data a)const{
		if(l == a.l) return r > a.r;
		return l < a.l;
	}
}d[233333];
 
 
//Seg_Tree
int ql, qr;
int t[999999];
int query(int now, int l, int r){
	if(l >= ql && r <= qr) return t[now];
	int mid = l + r >> 1, qa = INF, qb = INF;
	if(ql <= mid) qa = query(now << 1, l, mid);
	if(qr > mid) qb = query(now << 1 | 1, mid + 1, r);
	return MIN(qa, qb);
}
void modifm(int now, int l, int r, int q, int v){
	if(l == r) {t[now] = MIN(t[now], v); return;} 
	int mid = l + r >> 1;
	if(q <= mid) modifm(now << 1, l, mid, q, v);
	else modifm(now << 1 | 1, mid + 1, r, q, v);
	t[now] = MIN(t[now << 1], t[now << 1 | 1]);
}
 
int ans = 0;
void input(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		scanf("%d", &B[i]), (B[i] == 0) && (ans++);
	scanf("%d", &Q);
	for(int i = 1; i <= Q; i++)
		scanf("%d %d", &d[i].l, &d[i].r);
	std::sort(d + 1, d + Q + 1);
	memset(t, 63, sizeof(t));
}
 
void work(){
	int now = 1;
	for(int i = 1, tmp = 0; i <= n; i++){
		while(now <= Q && d[now].l == i){
			ql = i - 1, qr = d[now].r;
			tmp = query(1, 1, n);
			if(ql == 0) tmp = MIN(tmp, 0);
			modifm(1, 1, n, qr, tmp);
			now++;
		}
		ql = qr = i - 1;
		if(i == 1) tmp = 0;
		else tmp = query(1, 1, n);
		(B[i] == 0) ? (tmp--) : (tmp++);
		modifm(1, 1, n, i, tmp);
	}
	ql = qr = n;
	printf("%d\n", ans + query(1, 1, n));
}
 
int main(){
	input();
	work();
	return 0;
}
```