#include <cstdio>
#include <algorithm>
using namespace std;

typedef pair<int, int> pii;
const int MAXN = 2e5 + 10;
const int INF = 1e9;

int n;
int B[MAXN];
pii qs[MAXN];

int miseg[MAXN * 4], K;

void update(int i, int v) {
	i += K - 1;
	miseg[i] = v;
	i >>= 1;
	while (i > 0) {
		miseg[i] = min(miseg[2 * i], miseg[2 * i + 1]);
		i >>= 1;
	}
}

int query(int l, int r, int st, int ed, int idx) {
	if (r<st || l>ed) return INF;
	if (st <= l && r <= ed) return miseg[idx];

	int mid = (r + l) / 2;
	int lret = query(l, mid, st, ed, idx * 2);
	int rret = query(mid + 1, r, st, ed, idx * 2 + 1);
	return min(lret, rret);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &B[i]);

	int zcnt=0;
	for (int i = 1; i <= n; i++) {
		if(B[i]==0) zcnt++;
	}

	int q;
	scanf("%d", &q);
	for (int i = 0; i < q; i++) {
		int l, r;
		scanf("%d %d", &l, &r);
		qs[i] = { l, r };
	}
	sort(qs, qs + q);

	K = 1;
	while (K <= n) K <<= 1;
	for (int i = 2 * K - 1; i > 0; i--)
		miseg[i] = INF;

	int li = 1;
	int dis = 0;
	for (int i = 0; i < q; i++) {
		int l = qs[i].first, r = qs[i].second;
		while (li < l) {
			dis = min(dis + (B[li] == 1 ? 1 : -1), query(1, K, li, li, 1));
			// printf("idx:%d, dis:%d\n", li, dis);
			li++;
		}

		int mi = min(dis, query(1, K, l, r, 1));
		update(r, mi);
	}

	while (li <= n) {
		dis = min(dis + (B[li] == 1 ? 1 : -1), query(1, K, li, li, 1));
		// printf("idx:%d, dis:%d\n", li, dis);
		li++;
	}

	printf("%d", dis + zcnt);
	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 <algorithm>
using namespace std;

typedef pair<int, int> pii;
const int MAXN = 2e5 + 10;
const int INF = 1e9;

int n;
int B[MAXN];
pii qs[MAXN];

int miseg[MAXN * 4], K;

void update(int i, int v) {
	i += K - 1;
	miseg[i] = v;
	i >>= 1;
	while (i > 0) {
		miseg[i] = min(miseg[2 * i], miseg[2 * i + 1]);
		i >>= 1;
	}
}

int query(int l, int r, int st, int ed, int idx) {
	if (r<st || l>ed) return INF;
	if (st <= l && r <= ed) return miseg[idx];

	int mid = (r + l) / 2;
	int lret = query(l, mid, st, ed, idx * 2);
	int rret = query(mid + 1, r, st, ed, idx * 2 + 1);
	return min(lret, rret);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &B[i]);

	int zcnt=0;
	for (int i = 1; i <= n; i++) {
		if(B[i]==0) zcnt++;
	}

	int q;
	scanf("%d", &q);
	for (int i = 0; i < q; i++) {
		int l, r;
		scanf("%d %d", &l, &r);
		qs[i] = { l, r };
	}
	sort(qs, qs + q);

	K = 1;
	while (K <= n) K <<= 1;
	for (int i = 2 * K - 1; i > 0; i--)
		miseg[i] = INF;

	int li = 1;
	int dis = 0;
	for (int i = 0; i < q; i++) {
		int l = qs[i].first, r = qs[i].second;
		while (li < l) {
			dis = min(dis + (B[li] == 1 ? 1 : -1), query(1, K, li, li, 1));
			// printf("idx:%d, dis:%d\n", li, dis);
			li++;
		}

		int mi = min(dis, query(1, K, l, r, 1));
		update(r, mi);
	}

	while (li <= n) {
		dis = min(dis + (B[li] == 1 ? 1 : -1), query(1, K, li, li, 1));
		// printf("idx:%d, dis:%d\n", li, dis);
		li++;
	}

	printf("%d", dis + zcnt);
	return 0;
}
```