One day Petya got a set of wooden cubes as a present from his mom. Petya immediately built a whole city from these cubes.

The base of the city is an n × n square, divided into unit squares. The square's sides are parallel to the coordinate axes, the square's opposite corners have coordinates (0, 0) and (n, n). On each of the unit squares Petya built a tower of wooden cubes. The side of a wooden cube also has a unit length.

After that Petya went an infinitely large distance away from his masterpiece and looked at it in the direction of vector v = (vx, vy, 0). Petya wonders, how many distinct cubes are visible from this position. Help him, find this number.

Each cube includes the border. We think that a cube is visible if there is a ray emanating from some point p, belonging to the cube, in the direction of vector  - v, that doesn't contain any points, belonging to other cubes.

Input

The first line contains three integers n, vx and vy (1 ≤ n ≤ 103, |vx|, |vy| ≤ |104|, |vx| + |vy| > 0).

Next n lines contain n integers each: the j-th integer in the i-th line aij (0 ≤ aij ≤ 109, 1 ≤ i, j ≤ n) represents the height of the cube tower that stands on the unit square with opposite corners at points (i - 1, j - 1) and (i, j).

Output

Print a single integer — the number of visible cubes.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

5 -1 2
5 0 0 0 1
0 0 0 0 2
0 0 0 1 2
0 0 0 0 2
2 2 2 2 3


Output

20

Input

5 1 -2
5 0 0 0 1
0 0 0 0 2
0 0 0 1 2
0 0 0 0 2
2 2 2 2 3


Output

15
#include <bits/stdc++.h>
using namespace std;
struct node {
  long long mi, lzy;
} p[8020020];
long long n, x, y, hi[1005][1005], tot, chk[1002020], ans;
long long Q_y(long long i, long long j) { return x * j - y * i; }
void pushdn(long long i) {
  long long k = p[i].lzy;
  p[i << 1].mi = max(k, p[i << 1].mi);
  p[i << 1].lzy = max(k, p[i << 1].lzy);
  p[(i << 1) | 1].mi = max(k, p[(i << 1) | 1].mi);
  p[(i << 1) | 1].lzy = max(k, p[(i << 1) | 1].lzy);
  p[i].lzy = 0;
}
void in(long long x, long long y, long long k, long long l, long long r,
        long long i) {
  if (x <= l && r <= y) {
    p[i].mi = max(k, p[i].mi);
    p[i].lzy = max(p[i].lzy, k);
    return;
  }
  long long mid = (l + r) >> 1;
  if (p[i].lzy) pushdn(i);
  if (x <= mid) in(x, y, k, l, mid, i << 1);
  if (mid < y) in(x, y, k, mid + 1, r, (i << 1) | 1);
  p[i].mi = min(p[i << 1].mi, p[(i << 1) | 1].mi);
}
long long Q(long long x, long long y, long long l, long long r, long long i) {
  if (x <= l && r <= y) return p[i].mi;
  long long mid = (l + r) >> 1, s = 1000000009;
  if (p[i].lzy) pushdn(i);
  if (x <= mid) s = Q(x, y, l, mid, i << 1);
  if (mid < y) s = min(Q(x, y, mid + 1, r, (i << 1) | 1), s);
  return s;
}
int main() {
  scanf("%lld%lld%lld", &n, &x, &y);
  if (x >= 0) {
    if (y >= 0) {
      for (long long i = 1; i <= n; i++)
        for (long long j = 1; j <= n; j++) scanf("%lld", &hi[i][j]);
    } else {
      for (long long i = 1; i <= n; i++)
        for (long long j = n; j >= 1; j--) scanf("%lld", &hi[i][j]);
    }
  } else {
    if (y >= 0) {
      for (long long i = n; i >= 1; i--)
        for (long long j = 1; j <= n; j++) scanf("%lld", &hi[i][j]);
    } else {
      for (long long i = n; i >= 1; i--)
        for (long long j = n; j >= 1; j--) scanf("%lld", &hi[i][j]);
    }
  }
  x = abs(x);
  y = abs(y);
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= n; j++) chk[++tot] = Q_y(i, j);
  sort(chk + 1, chk + tot + 1);
  tot = unique(chk + 1, chk + tot + 1) - chk - 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++) {
      long long l = lower_bound(chk + 1, chk + tot + 1, Q_y(i, j - 1)) - chk,
                r = lower_bound(chk + 1, chk + tot + 1, Q_y(i - 1, j)) - chk,
                ss = Q(2 * l, 2 * r - 1, 1, 2 * tot, 1);
      if (hi[i][j] > ss) {
        ans += hi[i][j] - ss;
        in(2 * l, 2 * r - 1, hi[i][j], 1, 2 * tot, 1);
      }
    }
  printf("%lld", ans);
  return 0;
}
