Problem: Takahashi is a magician. He can cast a spell on an integer sequence (a_1,a_2,...,a_M) with M terms, to turn it into another sequence (s_1,s_2,...,s_M), where s_i is the sum of the first i terms in the original sequence.

One day, he received N integer sequences, each with M terms, and named those sequences A_1,A_2,...,A_N. He will try to cast the spell on those sequences so that A_1 < A_2 < ... < A_N will hold, where sequences are compared lexicographically. Let the action of casting the spell on a selected sequence be one cast of the spell. Find the minimum number of casts of the spell he needs to perform in order to achieve his objective.

Here, for two sequences a = (a_1,a_2,...,a_M), b = (b_1,b_2,...,b_M) with M terms each, a < b holds lexicographically if and only if there exists i (1 ≦ i ≦ M) such that a_j = b_j (1 ≦ j < i) and a_i < b_i.

Constraints

* 1 ≦ N ≦ 10^3
* 1 ≦ M ≦ 10^3
* Let the j-th term in A_i be A_{(i,j)}, then 1 ≦ A_{(i,j)} ≦ 10^9.

Input

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


N M
A_{(1,1)} A_{(1,2)} … A_{(1,M)}
A_{(2,1)} A_{(2,2)} … A_{(2,M)}
:
A_{(N,1)} A_{(N,2)} … A_{(N,M)}


Output

Print the minimum number of casts of the spell Takahashi needs to perform. If he cannot achieve his objective, print `-1` instead.

Examples

Input

3 3
2 3 1
2 1 2
2 6 3


Output

1


Input

3 3
3 2 10
10 5 4
9 1 9


Output

-1


Input

5 5
2 6 5 6 9
2 6 4 9 10
2 6 8 6 7
2 1 7 3 8
2 1 4 8 3


Output

11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m; cin >> n >> m;
	vector<vector<int>> a(n, vector<int>(m,0));
	for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) cin >> a[i][j];
	if(m == 1){
		for(int i = 0; i < n - 1; i++) if(a[i][0] >= a[i + 1][0]) return cout << -1 << endl, 0;
		return cout << 0 << endl, 0;
	}
	vector<long long> cost(n, 0ll), upto(n, 0ll);
	const long long sinf = 1<<30;
	auto ncr = [&](long long n, long long r){
		long long p = 1;
		for(int i = 1; i <= r; i++){
			p *= (n - r + i);
			p /= i;
			p = min(p, sinf);
		}
		return p;
	};
	auto less = [&](vector<int> a, vector<int> b, long long mvs, bool dfl){
		vector<long long> rb(b.begin(), b.end());
		const long long inf = 1ll<<60;
		if(mvs <= 15){
			for(int j = 0; j < mvs; j++){
				for(int i = 1; i < m; i++){
					rb[i] += rb[i - 1];
					rb[i] = min(rb[i], sinf);
				}
			}
		} else {
			vector<long long> to(m, 0ll);
			for(int i = 0; i < m; i++){
				if(to[i] == sinf) break;
				for(int j = i; j < m; j++){
					to[j] += b[i] * ncr(mvs - 1 + j - i, j - i);
					to[j] = min(to[j], sinf);
					if(to[j] == sinf) break;
				}
			}
			for(int j = 1; j < m; j++) to[j] = max(to[j], to[j - 1]);
			rb = to;
		}
		for(int i = 0; i < m; i++) if(a[i] != rb[i]){
			return a[i] < rb[i];
		}
		return dfl;
	};
	long long done = 0, ans = 0;
	for(int i = 0; i < n - 1; i++){
		if(a[i][0] > a[i + 1][0]) return cout << -1 << endl, 0;
		if(a[i][0] < a[i + 1][0]) {
			done = 0;
			continue;
		}
		int p = a[i][0], q = a[i][1];
		int x = a[i + 1][0], y = a[i + 1][1];
		if(p * done + q < y) done = 0;
		else {
			long long lo = 0, hi = 1ll<<32, ans = 0;
			while(lo <= hi){
				long long mid = (lo + hi)>>1;
				if(x * mid + y <= p * done + q) ans = mid, lo = mid + 1;
				else hi = mid - 1;
			}
			while(true){
				long long b = ans;
				if(done < ans){
					if(!less(a[i], a[i + 1], ans - done, false)) ans++;
				} else {
					if(less(a[i + 1], a[i], done - ans, true)) ans++;
				}
				if(b == ans) break;
			}
			done = ans;
		}
		ans += done;
	}
	cout << ans << endl;
	return 0;
}