We have a weighted directed graph with N vertices numbered 0 to N-1.

The graph initially has N-1 edges. The i-th edge (0 \leq i \leq N-2) is directed from Vertex i to Vertex i+1 and has a weight of 0.

Snuke will now add a new edge (i → j) for every pair i, j (0 \leq i,j \leq N-1,\ i \neq j). The weight of the edge will be -1 if i < j, and 1 otherwise.

Ringo is a boy. A negative cycle (a cycle whose total weight is negative) in a graph makes him sad. He will delete some of the edges added by Snuke so that the graph will no longer contain a negative cycle. The cost of deleting the edge (i → j) is A_{i,j}. He cannot delete edges that have been present from the beginning.

Find the minimum total cost required to achieve Ringo's objective.

Constraints

* 3 \leq N \leq 500
* 1 \leq A_{i,j} \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_{0,1} A_{0,2} A_{0,3} \cdots A_{0,N-1}
A_{1,0} A_{1,2} A_{1,3} \cdots A_{1,N-1}
A_{2,0} A_{2,1} A_{2,3} \cdots A_{2,N-1}
\vdots
A_{N-1,0} A_{N-1,1} A_{N-1,2} \cdots A_{N-1,N-2}


Output

Print the minimum total cost required to achieve Ringo's objective.

Examples

Input

3
2 1
1 4
3 3


Output

2


Input

4
1 1 1
1 1 1
1 1 1
1 1 1


Output

2


Input

10
190587 2038070 142162180 88207341 215145790 38 2 5 20
32047998 21426 4177178 52 734621629 2596 102224223 5 1864
41 481241221 1518272 51 772 146 8805349 3243297 449
918151 126080576 5186563 46354 6646 491776 5750138 2897 161
3656 7551068 2919714 43035419 495 3408 26 3317 2698
455357 3 12 1857 5459 7870 4123856 2402 258
3 25700 16191 102120 971821039 52375 40449 20548149 16186673
2 16 130300357 18 6574485 29175 179 1693 2681
99 833 131 2 414045824 57357 56 302669472 95
8408 7 1266941 60620177 129747 41382505 38966 187 5151064


Output

2280211
#include<bits/stdc++.h>
using namespace std;

// the distance functions are in blocks
// forward edges only to future blocks
// backwards edges only to self & immediate predecessor

using ll = long long;

const int MAXN = 600;
int N;
ll A[MAXN][MAXN];

ll backPref[MAXN][MAXN];
ll forwPref[MAXN][MAXN];

ll dp[MAXN][MAXN]; // max value we can save

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (j == i) A[i][j] = 0;
			else cin >> A[i][j];
		}
	}

	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (i == j) continue;
			if (i < j) {
				forwPref[i+1][j+1] = A[i][j];
			} else {
				backPref[i+1][j+1] = A[i][j];
			}
		}
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			forwPref[i+1][j+1] += forwPref[i][j+1] + forwPref[i+1][j] - forwPref[i][j];
			backPref[i+1][j+1] += backPref[i][j+1] + backPref[i+1][j] - backPref[i][j];
		}
	}

	for (int j = 1; j < N; j++) {
		// back [0,j) by [0,j)
		// no forwards
		dp[0][j] = backPref[j][j];
	}
	for (int i = 1; i < N; i++) {
		for (int j = i+1; j <= N; j++) {
			for (int k = 0; k < i; k++) {
				// back [i,j) by [k,j)
				// forwards [0,i) by [i,j)
				dp[i][j] = max(dp[i][j], dp[k][i]
						+ (backPref[j][j] - backPref[i][j] - backPref[j][k] + backPref[i][k])
						+ (forwPref[i][j] - forwPref[0][j] - forwPref[i][i] + forwPref[0][i])
						);
			}
		}
	}

	ll ans = 0;
	for (int i = 0; i < N; i++) {
		ans = max(ans, dp[i][N]);
	}

	ans = -ans;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (i == j) continue;
			ans += A[i][j];
		}
	}
	cout << ans << '\n';

	return 0;
}
