Problem: Nuske has a grid with N rows and M columns of squares. The rows are numbered 1 through N from top to bottom, and the columns are numbered 1 through M from left to right. Each square in the grid is painted in either blue or white. If S_{i,j} is 1, the square at the i-th row and j-th column is blue; if S_{i,j} is 0, the square is white. For every pair of two blue square a and b, there is at most one path that starts from a, repeatedly proceeds to an adjacent (side by side) blue square and finally reaches b, without traversing the same square more than once.

Phantom Thnook, Nuske's eternal rival, gives Q queries to Nuske. The i-th query consists of four integers x_{i,1}, y_{i,1}, x_{i,2} and y_{i,2} and asks him the following: when the rectangular region of the grid bounded by (and including) the x_{i,1}-th row, x_{i,2}-th row, y_{i,1}-th column and y_{i,2}-th column is cut out, how many connected components consisting of blue squares there are in the region?

Process all the queries.

Constraints

* 1 ≤ N,M ≤ 2000
* 1 ≤ Q ≤ 200000
* S_{i,j} is either 0 or 1.
* S_{i,j} satisfies the condition explained in the statement.
* 1 ≤ x_{i,1} ≤ x_{i,2} ≤ N(1 ≤ i ≤ Q)
* 1 ≤ y_{i,1} ≤ y_{i,2} ≤ M(1 ≤ i ≤ Q)

Input

The input is given from Standard Input in the following format:


N M Q
S_{1,1}..S_{1,M}
:
S_{N,1}..S_{N,M}
x_{1,1} y_{i,1} x_{i,2} y_{i,2}
:
x_{Q,1} y_{Q,1} x_{Q,2} y_{Q,2}


Output

For each query, print the number of the connected components consisting of blue squares in the region.

Examples

Input

3 4 4
1101
0110
1101
1 1 3 4
1 1 3 1
2 2 3 4
1 2 2 4


Output

3
2
2
2


Input

5 5 6
11010
01110
10101
11101
01010
1 1 5 5
1 2 4 5
2 3 3 4
3 3 3 3
3 1 3 5
1 1 3 4


Output

3
2
1
1
3
2

Solution (in JAVA):

Solution:
import static java.lang.Math.*;
import static java.util.Arrays.*;
import java.util.*;
import java.io.*;

public class Main {

	void solve() {
		int R = sc.nextInt();
		int C = sc.nextInt();
		int Q = sc.nextInt();

		char[][] S = new char[R][];
		for (int r = 0; r < R; r++) {
			S[r] = sc.next().toCharArray();
		}
		
		int[][] vs = new int[R+1][C+1];
		int[][] e0 = new int[R+1][C+1];
		int[][] e1 = new int[R+1][C+1];
		
		for (int r = 0; r < R; r++) {
			for (int c = 0; c < C; c++) {
				vs[r+1][c+1] = vs[r][c+1] + vs[r+1][c] - vs[r][c] + (S[r][c] =='1' ? 1 : 0);
			}
		}
		
		for (int r = 0; r < R; r++) {
			for (int c = 0; c + 1 < C; c++) {
				e0[r+1][c+1] = e0[r][c+1] + e0[r+1][c] - e0[r][c] + 
					(S[r][c] =='1' && S[r][c+1] =='1' ? 1 : 0);
			}
		}

		for (int r = 0; r + 1 < R; r++) {
			for (int c = 0; c < C; c++) {
				e1[r+1][c+1] = e1[r][c+1] + e1[r+1][c] - e1[r][c] + 
					(S[r][c] =='1' && S[r+1][c] == '1' ? 1 : 0);
			}
		}

		for (int q = 0; q < Q; q++) {
			int x1 = sc.nextInt() - 1;
			int y1 = sc.nextInt() - 1;
			int x2 = sc.nextInt();
			int y2 = sc.nextInt();
			int V = vs[x2][y2] - vs[x2][y1] - vs[x1][y2] + vs[x1][y1];
			int E0 = e0[x2][y2-1] - e0[x2][y1] - e0[x1][y2-1] + e0[x1][y1];
			int E1 = e1[x2-1][y2] - e1[x2-1][y1] - e1[x1][y2] + e1[x1][y1];
			tr(V, E0, E1);
			int ans = V - (E0 + E1);
			out.println(ans);
		}

	}

	static void tr(Object... os) { System.err.println(deepToString(os)); }

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

	MyScanner sc = null;
	PrintWriter out = null;
	public void run() throws Exception {
		sc = new MyScanner(System.in);
		out = new PrintWriter(System.out);
		for (;sc.hasNext();) {
			solve();
			out.flush();
		}
		out.close();
	}

	class MyScanner {
		String line;
		BufferedReader reader;
		StringTokenizer tokenizer;

		public MyScanner(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}
		public void eat() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					line = reader.readLine();
					if (line == null) {
						tokenizer = null;
						return;
					}
					tokenizer = new StringTokenizer(line);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}
		public String next() {
			eat();
			return tokenizer.nextToken();
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		public boolean hasNext() {
			eat();
			return (tokenizer != null && tokenizer.hasMoreElements());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
	}
}