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

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		new Main().run();
	}

	void run() {
		Scanner sc = new Scanner(System.in);
		int base = 0;
		int n = sc.nextInt();
		int[] b = new int[n];
		for (int i = 0; i < n; ++i) {
			b[i] = sc.nextInt();
			if (b[i] == 0)
				++base;
		}
		int Q = sc.nextInt();
		int[][] q = new int[Q][2];
		for (int i = 0; i < Q; ++i) {
			q[i][0] = sc.nextInt();
			q[i][1] = sc.nextInt();
		}
		Arrays.sort(q, new Comparator<int[]>() {
			@Override
			public int compare(int[] o1, int[] o2) {
				return Integer.compare(o1[0], o2[0]);
			}
		});
		RMQ rmq = new RMQ(n + 1);
		rmq.set(0, 0);
		int p = 0;
		for (int i = 0; i < Q; ++i) {
			while (q[i][0] - p > 1) {
				int c = b[p] == 1 ? 1 : -1;
				rmq.set(p + 1, Math.min(c + rmq.query(p, p + 1), rmq.query(p + 1, p + 2)));
				++p;
			}
			rmq.set(q[i][1], rmq.query(p, q[i][1] + 1));
		}
		while (n - p > 0) {
			int c = b[p] == 1 ? 1 : -1;
			rmq.set(p + 1, Math.min(c + rmq.query(p, p + 1), rmq.query(p + 1, p + 2)));
			++p;
		}
		System.out.println(base + rmq.query(n, n + 1));
	}

	class RMQ {
		int n;
		int[] v;

		public RMQ(int n_) {
			n = 1;
			while (n < n_)
				n *= 2;
			v = new int[2 * n - 1];
			Arrays.fill(v, Integer.MAX_VALUE / 3);
		}

		void add(int k, int val) {
			k += n - 1;
			v[k] += val;
			while (k > 0) {
				k = (k - 1) / 2;
				v[k] = Math.min(v[2 * k + 1], v[2 * k + 2]);
			}
		}

		void set(int k, int val) {
			k += n - 1;
			v[k] = val;
			while (k > 0) {
				k = (k - 1) / 2;
				v[k] = Math.min(v[2 * k + 1], v[2 * k + 2]);
			}
		}

		int query(int a, int b) {
			return query(a, b, 0, n, 0);
		}

		int query(int a, int b, int l, int r, int k) {
			if (r <= a || b <= l)
				return Integer.MAX_VALUE / 3;
			else if (a <= l && r <= b)
				return v[k];
			else
				return Math.min(query(a, b, l, (l + r) / 2, 2 * k + 1), query(a, b, (l + r) / 2, r, 2 * k + 2));
		}
	}

	void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}