If there is an integer not less than 0 satisfying the following conditions, print the smallest such integer; otherwise, print `-1`.

* The integer has exactly N digits in base ten. (We assume 0 to be a 1-digit integer. For other integers, leading zeros are not allowed.)
* The s_i-th digit from the left is c_i. \left(i = 1, 2, \cdots, M\right)

Constraints

* All values in input are integers.
* 1 \leq N \leq 3
* 0 \leq M \leq 5
* 1 \leq s_i \leq N
* 0 \leq c_i \leq 9

Input

Input is given from Standard Input in the following format:


N M
s_1 c_1
\vdots
s_M c_M


Output

Print the answer.

Examples

Input

3 3
1 7
3 2
1 7


Output

702


Input

3 2
2 1
2 3


Output

-1


Input

3 1
1 0


Output

-1
#include<bits/stdc++.h>
using namespace std;
int main() {
	int n, m;cin >> n >> m;
	vector<int> d(n, -1);
	bool b = true;
	for (int i = 0; i < m; ++i) {
		int s, c;cin >> s >> c;
		if (d[s-1] != -1 && d[s-1] != c) b = false;
		else if (s == 1 && c == 0 && n > 1) b = false;
		else d[s-1] = c;
	}
	if (!b) {
		cout << "-1\n";
		exit(0);
	}
	if (d[0] == -1 && d.size() > 1) d[0] = 1;
	for (int i = 0; i < n; ++i) 
		if (d[i] == -1) d[i] = 0;
	for (int u : d) cout << u;
}