#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#define Root 0, n, 1
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef long long LL;

inline int read() {
	int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005;
int T[N << 2], b[N];
vector<int> vec[N];

bool cmp(int x,int y) { return x > y; }
void update(int l,int r,int rt,int p,int v) {
	if (l == r) { T[rt] = min(T[rt], v); return ; }
	int mid = (l + r) >> 1;
	if (p <= mid) update(lson, p, v);
	else update(rson, p, v);
	T[rt] = min(T[rt << 1], T[rt << 1 | 1]);	
}
int query(int l,int r,int rt,int L,int R) {
	if (L <= l && r <= R) return T[rt];
	int mid = (l + r) >> 1, res = 1e9;
	if (L <= mid) res = min(res, query(lson, L, R));
	if (R > mid) res = min(res, query(rson, L, R));
	return res;
}
int main() {
	int n = read(), sum = 0;
	for (int i = 1; i <= n; ++i) {
		b[i] = read(); sum += (b[i] == 0);
	}
	int m = read();
	for (int i = 1; i <= m; ++i) {
		int l = read(), r = read(); vec[l].push_back(r);
	}
	for (int i = 1; i <= n; ++i) sort(vec[i].begin(), vec[i].end(), cmp);
	memset(T, 0x3f, sizeof(T));
	update(Root, 0, 0);
	for (int i = 1; i <= n; ++i) {
		for (int sz = vec[i].size(), j = 0; j < sz; ++j) {
			int t = query(Root, i - 1, vec[i][j]);
			update(Root, vec[i][j], t);
		}
		int t = query(Root, i - 1, i - 1);
		update(Root, i, t + (b[i] ? 1 : -1));
	}
	cout << sum + query(Root, n, n);
	return 0;
}
/*

5
0 1 0 1 0
1
1 5

*/
### Prompt

In Cpp, your task is to solve 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<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#define Root 0, n, 1
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef long long LL;

inline int read() {
	int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005;
int T[N << 2], b[N];
vector<int> vec[N];

bool cmp(int x,int y) { return x > y; }
void update(int l,int r,int rt,int p,int v) {
	if (l == r) { T[rt] = min(T[rt], v); return ; }
	int mid = (l + r) >> 1;
	if (p <= mid) update(lson, p, v);
	else update(rson, p, v);
	T[rt] = min(T[rt << 1], T[rt << 1 | 1]);	
}
int query(int l,int r,int rt,int L,int R) {
	if (L <= l && r <= R) return T[rt];
	int mid = (l + r) >> 1, res = 1e9;
	if (L <= mid) res = min(res, query(lson, L, R));
	if (R > mid) res = min(res, query(rson, L, R));
	return res;
}
int main() {
	int n = read(), sum = 0;
	for (int i = 1; i <= n; ++i) {
		b[i] = read(); sum += (b[i] == 0);
	}
	int m = read();
	for (int i = 1; i <= m; ++i) {
		int l = read(), r = read(); vec[l].push_back(r);
	}
	for (int i = 1; i <= n; ++i) sort(vec[i].begin(), vec[i].end(), cmp);
	memset(T, 0x3f, sizeof(T));
	update(Root, 0, 0);
	for (int i = 1; i <= n; ++i) {
		for (int sz = vec[i].size(), j = 0; j < sz; ++j) {
			int t = query(Root, i - 1, vec[i][j]);
			update(Root, vec[i][j], t);
		}
		int t = query(Root, i - 1, i - 1);
		update(Root, i, t + (b[i] ? 1 : -1));
	}
	cout << sum + query(Root, n, n);
	return 0;
}
/*

5
0 1 0 1 0
1
1 5

*/
```