#include <stdio.h>
#include <string.h>
#include <vector>
int n, b[200005], dp[200005], cnt, M[200005 << 2], q, x, y;
std::vector<int> v[200005];
void modify(int cur, int l, int r, int pos, int num) {
	M[cur] = std::min(M[cur], num);
	if (l == r) return;
	int mid = l + r >> 1;
	if (pos <= mid) modify(cur << 1, l, mid, pos, num);
	else modify(cur << 1 | 1, mid + 1, r, pos, num);
}
int query(int cur, int l, int r, int L, int R) {
	if (L <= l && R >= r) return M[cur];
	int mid = l + r >> 1, ans = 0x3f3f3f3f;
	if (L <= mid) ans = std::min(ans, query(cur << 1, l, mid, L, R));
	if (R > mid) ans = std::min(ans, query(cur << 1 | 1, mid + 1, r, L, R));
	return ans;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &b[i]), cnt += b[i] == 0;
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) scanf("%d%d", &x, &y), v[x].push_back(y);
	memset(M, 0x3f, sizeof M); memset(dp, 0x3f, sizeof dp);
	dp[0] = 0; modify(1, 0, n, 0, 0);
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < v[i].size(); j++) 
			modify(1, 0, n, v[i][j], dp[v[i][j]] = std::min(dp[v[i][j]], query(1, 0, n, i - 1, v[i][j])));
		modify(1, 0, n, i, dp[i] = std::min(dp[i], dp[i - 1] + (b[i] ? 1 : -1)));
	}
	printf("%d\n", dp[n] + cnt);
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <stdio.h>
#include <string.h>
#include <vector>
int n, b[200005], dp[200005], cnt, M[200005 << 2], q, x, y;
std::vector<int> v[200005];
void modify(int cur, int l, int r, int pos, int num) {
	M[cur] = std::min(M[cur], num);
	if (l == r) return;
	int mid = l + r >> 1;
	if (pos <= mid) modify(cur << 1, l, mid, pos, num);
	else modify(cur << 1 | 1, mid + 1, r, pos, num);
}
int query(int cur, int l, int r, int L, int R) {
	if (L <= l && R >= r) return M[cur];
	int mid = l + r >> 1, ans = 0x3f3f3f3f;
	if (L <= mid) ans = std::min(ans, query(cur << 1, l, mid, L, R));
	if (R > mid) ans = std::min(ans, query(cur << 1 | 1, mid + 1, r, L, R));
	return ans;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &b[i]), cnt += b[i] == 0;
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) scanf("%d%d", &x, &y), v[x].push_back(y);
	memset(M, 0x3f, sizeof M); memset(dp, 0x3f, sizeof dp);
	dp[0] = 0; modify(1, 0, n, 0, 0);
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < v[i].size(); j++) 
			modify(1, 0, n, v[i][j], dp[v[i][j]] = std::min(dp[v[i][j]], query(1, 0, n, i - 1, v[i][j])));
		modify(1, 0, n, i, dp[i] = std::min(dp[i], dp[i - 1] + (b[i] ? 1 : -1)));
	}
	printf("%d\n", dp[n] + cnt);
	return 0;
}
```