Fox Ciel has a board with n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol '.', or a symbol '#'.

A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.

<image>

Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.

Please, tell Ciel if she can draw the crosses in the described way.

Input

The first line contains an integer n (3 ≤ n ≤ 100) — the size of the board.

Each of the next n lines describes one row of the board. The i-th line describes the i-th row of the board and consists of n characters. Each character is either a symbol '.', or a symbol '#'.

Output

Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".

Examples

Input

5
.#...
####.
.####
...#.
.....


Output

YES


Input

4
####
####
####
####


Output

NO


Input

6
.#....
####..
.####.
.#.##.
######
.#..#.


Output

YES


Input

6
.#..#.
######
.####.
.####.
######
.#..#.


Output

NO


Input

3
...
...
...


Output

YES

Note

In example 1, you can draw two crosses. The picture below shows what they look like.

<image>

In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.
#include <bits/stdc++.h>
using namespace std;
template <class T>
string toString(T n) {
  ostringstream ost;
  ost << n;
  ost.flush();
  return ost.str();
}
int toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
long long toLInt(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
vector<string> parse(string temp) {
  vector<string> ans;
  ans.clear();
  string s;
  istringstream iss(temp);
  while (iss >> s) ans.push_back(s);
  return ans;
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <class T>
inline T power(T b, T p) {
  if (p < 0) return -1;
  if (b <= 0) return -2;
  if (!p) return 1;
  return b * power(b, p - 1);
}
template <class T>
int bit_count(T no) {
  int cnt = 0;
  while (no) {
    no = no >> 1;
    cnt++;
  }
  return cnt;
}
void seive(int lm, int arr[]) {
  for (int i = 2; i * i <= lm; i++)
    if (arr[i] == 0)
      for (int j = 1; j * i <= lm; j++) arr[i * j] = 1;
}
template <class T>
void showBit(T N) {
  T zero = 0, one = 1;
  if (N == zero) return;
  showBit(N >> 1);
  printf("%d", N & one);
}
long long binToDec(string s) {
  int ln = s.length();
  long long sum = 0;
  for (int i = ln - 1, k = 0; i >= 0; i--, k++)
    sum = sum + (s[i] - '0') * (1 << k);
  return sum;
}
void combination(long long com[][1005]) {
  com[0][0] = 1;
  int sz = 1000;
  for (int i = 1; i <= sz; i++) {
    com[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      com[i][j] = com[i - 1][j] + com[i - 1][j - 1];
      com[i][j] = com[i][j] % 1000000007;
    }
  }
}
template <class T>
T bigmod(T bs, T p, T mod) {
  if (mod <= 0) {
    printf("Exception");
    return 0;
  }
  T ans = 1;
  T k = bs;
  while (p) {
    if (p & 1) {
      ans = ans * k;
      ans %= mod;
    }
    k *= k;
    k %= mod;
    p = p >> 1;
  }
  return ans;
}
void insert_ind(int cs, int ci, int s, int e) {
  if (s == e) {
    return;
  }
  int mid = (s + e) / 2;
  if (ci <= mid)
    insert_ind(2 * cs, ci, s, mid);
  else
    insert_ind(2 * cs + 1, ci, mid + 1, e);
}
void update_ind(int cs, int ci, int s, int e, int val) {
  if (s == e) {
    return;
  }
  int mid = (s + e) / 2;
  if (ci <= mid)
    update_ind(2 * cs, ci, s, mid, val);
  else
    update_ind(2 * cs + 1, ci, mid + 1, e, val);
}
void update_seg(int cs, int ds, int de, int s, int e, int val) {
  if (ds == s && de == e) {
    return;
  }
  int mid = (s + e) / 2;
  if (ds <= mid) update_seg(2 * cs, ds, min(mid, de), s, mid, val);
  if (de > mid) update_seg(2 * cs + 1, max(mid + 1, ds), de, mid + 1, e, val);
}
void pattern_match(char pat[], const int len, int pi[]) {
  pi[0] = 0;
  for (int i = 1; i < len; i++) {
    int ch = pi[i - 1];
    while (pat[ch] != pat[i]) {
      if (ch == 0) {
        ch = -1;
        break;
      }
      ch = pi[ch - 1];
    }
    pi[i] = ch + 1;
  }
}
void pattern_match(char pat[], int pi[], char tar[], const int len1, int gi[]) {
  if (pat[0] == tar[0]) gi[0] = 1;
  for (int i = 1; i < len1; i++) {
    int st = gi[i - 1];
    while (pat[st] != tar[i]) {
      if (st == 0) {
        st = -1;
        break;
      }
      st = pi[st - 1];
    }
    if (st == -1)
      gi[i] = 0;
    else
      gi[i] = st + 1;
  }
}
struct myArray {
  int mat[10][10];
};
myArray matrix_mul(myArray mat1, myArray mat2, const int sz) {
  int tmp[sz][sz];
  memset((tmp), 0, sizeof((tmp)));
  for (int r = 0; r < sz; r++)
    for (int c = 0; c < sz; c++)
      for (int cr = 0; cr < sz; cr++)
        tmp[r][c] += mat1.mat[r][cr] * mat2.mat[cr][c];
  myArray res;
  for (int r = 0; r < sz; r++)
    for (int c = 0; c < sz; c++) res.mat[r][c] = tmp[r][c];
  return res;
}
myArray unit_matrix(const int sz) {
  myArray ma;
  memset((ma.mat), 0, sizeof((ma.mat)));
  for (int i = 0; i < sz; i++) ma.mat[i][i] = 1;
  return ma;
}
myArray matrix_expo(int p, myArray mat, const int sz) {
  if (p == 0) {
    return unit_matrix(sz);
  }
  if (p & 1) return matrix_mul(mat, matrix_expo(p - 1, mat, sz), sz);
  myArray tmp = matrix_expo(p >> 1, mat, sz);
  return matrix_mul(tmp, tmp, sz);
}
map<string, int> msi;
void cal_state(string s, int tran[][30]) {
  int ln = s.length();
  string s1 = s;
  for (int i = 0; i < ln; i++) {
    msi[s1] = ln - i;
    s1 = s1.substr(0, ln - i - 1);
  }
  s1 = "";
  for (int i = 0; i < ln; i++) {
    string ts = s1;
    for (int j = 0; j < 26; j++) {
      s1 = ts + (char)('a' + j);
      int tln = i + 1;
      while (msi[s1] == 0) {
        s1 = s1.substr(1);
        tln--;
        if (tln == 0) break;
      }
      if (tln == 0) s1 = "";
      tran[i][j] = msi[s1];
    }
    s1 = s.substr(0, i + 1);
  }
}
double distance(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int X[] = {0, 0, 1, -1};
int Y[] = {1, -1, 0, 0};
const int dx[8] = {-1, 1, -2, 2, -1, 1, -2, 2};
const int dy[8] = {2, 2, -1, -1, -2, -2, 1, 1};
int n, cnt, ci, cj;
string st[105];
bool valid(int i) {
  if (i >= 0 && i < n) return true;
  return false;
}
bool has(int i, int j) {
  if (valid(i) && valid(j) && st[i][j] == '#') return true;
  return false;
}
bool hasCross() {
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (n); ++j)
      if (st[i][j] == '#') {
        if (has(i, j) && has(i + 1, j) && has(i + 2, j) && has(i + 1, j - 1) &&
            has(i + 1, j + 1)) {
          ci = i, cj = j;
          return true;
        }
        return false;
      }
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) cin >> st[i];
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (n); ++j)
      if (st[i][j] == '#') cnt++;
  if (cnt == 0) {
    cout << "YES\n";
    return 0;
  }
  while (cnt) {
    bool is = hasCross();
    if (!is) {
      cout << "NO\n";
      return 0;
    }
    cnt -= 5;
    st[ci][cj] = st[ci + 1][cj] = st[ci + 2][cj] = st[ci + 1][cj - 1] =
        st[ci + 1][cj + 1] = '.';
  }
  cout << "YES\n";
  return 0;
}
