#include <bits/stdc++.h>
using namespace std;
int mat[1001][1001][2];
int dp1[1001][1001];
int dp2[1001][1001];
int dx[2] = {0, 1};
int dy[2] = {1, 0};
vector<pair<int, int> > vx;
void findxx1(int i, int j);
void findxx2(int i, int j);
int chk(int x, int y, int n) {
  if (x >= n || y >= n || y < 0 || x < 0)
    return 0;
  else
    return 1;
}
struct parent {
  int x;
  int y;
};
struct parent par1[1001][1001], par2[1001][1001];
struct node {
  int x;
  int y;
};
void bfs2(int n) {
  int i, j, x, y;
  queue<node> q;
  node a;
  a.x = 0;
  a.y = 0;
  q.push(a);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) dp1[i][j] = 99999;
  dp1[0][0] = mat[0][0][0];
  while (!q.empty()) {
    x = q.front().x;
    y = q.front().y;
    q.pop();
    for (i = 0; i < 2; i++) {
      int x1, y1;
      x1 = x + dx[i];
      y1 = y + dy[i];
      if (chk(x1, y1, n)) {
        if (dp1[x1][y1] > dp1[x][y] + mat[x1][y1][0]) {
          dp1[x1][y1] = dp1[x][y] + mat[x1][y1][0];
          a.x = x1;
          a.y = y1;
          q.push(a);
          par1[x1][y1].x = x;
          par1[x1][y1].y = y;
        }
      }
    }
  }
}
void bfs5(int n) {
  int i, j, x, y;
  queue<node> q;
  node a;
  a.x = 0;
  a.y = 0;
  q.push(a);
  for (int i = 0; i < n; i++)
    for (j = 0; j < n; j++) dp2[i][j] = 99999;
  dp2[0][0] = mat[0][0][1];
  while (!q.empty()) {
    x = q.front().x;
    y = q.front().y;
    q.pop();
    for (i = 0; i < 2; i++) {
      int x1, y1;
      x1 = x + dx[i];
      y1 = y + dy[i];
      if (chk(x1, y1, n)) {
        if (dp2[x1][y1] > dp2[x][y] + mat[x1][y1][1]) {
          dp2[x1][y1] = dp2[x][y] + mat[x1][y1][1];
          a.x = x1;
          a.y = y1;
          q.push(a);
          par2[x1][y1].x = x;
          par2[x1][y1].y = y;
        }
      }
    }
  }
}
void zeroPrint(int x, int y, int n) {
  int i, j, k;
  for (i = 0; i < x; i++) cout << "D";
  for (i = 0; i < y; i++) cout << "R";
  for (i = x; i < n - 1; i++) cout << "D";
  for (i = y; i < n - 1; i++) cout << "R";
}
int main() {
  ios_base::sync_with_stdio(0);
  vx.reserve(10000);
  int n, i, j, k, x, xx, yy, zero = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> x;
      if (x == 0) {
        zero = 1;
        xx = i;
        yy = j;
      } else {
        int tmp = x;
        while (tmp && tmp % 2 == 0) {
          mat[i][j][0]++;
          tmp /= 2;
        }
        while (tmp && tmp % 5 == 0) {
          mat[i][j][1]++;
          tmp /= 5;
        }
      }
    }
  }
  bfs2(n);
  bfs5(n);
  int ret = min(dp1[n - 1][n - 1], dp2[n - 1][n - 1]);
  if (zero && ret > 1) {
    cout << 1 << endl;
    zeroPrint(xx, yy, n);
  } else {
    cout << ret << endl;
    vx.push_back({n - 1, n - 1});
    if (dp1[n - 1][n - 1] > dp2[n - 1][n - 1]) {
      findxx2(n - 1, n - 1);
    } else
      findxx1(n - 1, n - 1);
    reverse(vx.begin(), vx.end());
    for (i = 1; i < vx.size(); i++) {
      if (vx[i - 1].first == vx[i].first)
        cout << "R";
      else
        cout << "D";
    }
    cout << endl;
  }
}
void findxx1(int i, int j) {
  if (i == 0 && j == 0) {
    return;
  } else {
    vx.push_back({par1[i][j].x, par1[i][j].y});
    findxx1(par1[i][j].x, par1[i][j].y);
  }
}
void findxx2(int i, int j) {
  if (i == 0 && j == 0) {
    return;
  } else {
    vx.push_back({par2[i][j].x, par2[i][j].y});
    findxx2(par2[i][j].x, par2[i][j].y);
  }
}

### Prompt

Generate a cpp solution to the following problem:

There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that 

  * starts in the upper left cell of the matrix; 
  * each following cell is to the right or down from the current cell; 
  * the way ends in the bottom right cell. 



Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.

Input

The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 109).

Output

In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.

Examples

Input

3
1 2 3
4 5 6
7 8 9


Output

0
DDRR

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int mat[1001][1001][2];
int dp1[1001][1001];
int dp2[1001][1001];
int dx[2] = {0, 1};
int dy[2] = {1, 0};
vector<pair<int, int> > vx;
void findxx1(int i, int j);
void findxx2(int i, int j);
int chk(int x, int y, int n) {
  if (x >= n || y >= n || y < 0 || x < 0)
    return 0;
  else
    return 1;
}
struct parent {
  int x;
  int y;
};
struct parent par1[1001][1001], par2[1001][1001];
struct node {
  int x;
  int y;
};
void bfs2(int n) {
  int i, j, x, y;
  queue<node> q;
  node a;
  a.x = 0;
  a.y = 0;
  q.push(a);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) dp1[i][j] = 99999;
  dp1[0][0] = mat[0][0][0];
  while (!q.empty()) {
    x = q.front().x;
    y = q.front().y;
    q.pop();
    for (i = 0; i < 2; i++) {
      int x1, y1;
      x1 = x + dx[i];
      y1 = y + dy[i];
      if (chk(x1, y1, n)) {
        if (dp1[x1][y1] > dp1[x][y] + mat[x1][y1][0]) {
          dp1[x1][y1] = dp1[x][y] + mat[x1][y1][0];
          a.x = x1;
          a.y = y1;
          q.push(a);
          par1[x1][y1].x = x;
          par1[x1][y1].y = y;
        }
      }
    }
  }
}
void bfs5(int n) {
  int i, j, x, y;
  queue<node> q;
  node a;
  a.x = 0;
  a.y = 0;
  q.push(a);
  for (int i = 0; i < n; i++)
    for (j = 0; j < n; j++) dp2[i][j] = 99999;
  dp2[0][0] = mat[0][0][1];
  while (!q.empty()) {
    x = q.front().x;
    y = q.front().y;
    q.pop();
    for (i = 0; i < 2; i++) {
      int x1, y1;
      x1 = x + dx[i];
      y1 = y + dy[i];
      if (chk(x1, y1, n)) {
        if (dp2[x1][y1] > dp2[x][y] + mat[x1][y1][1]) {
          dp2[x1][y1] = dp2[x][y] + mat[x1][y1][1];
          a.x = x1;
          a.y = y1;
          q.push(a);
          par2[x1][y1].x = x;
          par2[x1][y1].y = y;
        }
      }
    }
  }
}
void zeroPrint(int x, int y, int n) {
  int i, j, k;
  for (i = 0; i < x; i++) cout << "D";
  for (i = 0; i < y; i++) cout << "R";
  for (i = x; i < n - 1; i++) cout << "D";
  for (i = y; i < n - 1; i++) cout << "R";
}
int main() {
  ios_base::sync_with_stdio(0);
  vx.reserve(10000);
  int n, i, j, k, x, xx, yy, zero = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> x;
      if (x == 0) {
        zero = 1;
        xx = i;
        yy = j;
      } else {
        int tmp = x;
        while (tmp && tmp % 2 == 0) {
          mat[i][j][0]++;
          tmp /= 2;
        }
        while (tmp && tmp % 5 == 0) {
          mat[i][j][1]++;
          tmp /= 5;
        }
      }
    }
  }
  bfs2(n);
  bfs5(n);
  int ret = min(dp1[n - 1][n - 1], dp2[n - 1][n - 1]);
  if (zero && ret > 1) {
    cout << 1 << endl;
    zeroPrint(xx, yy, n);
  } else {
    cout << ret << endl;
    vx.push_back({n - 1, n - 1});
    if (dp1[n - 1][n - 1] > dp2[n - 1][n - 1]) {
      findxx2(n - 1, n - 1);
    } else
      findxx1(n - 1, n - 1);
    reverse(vx.begin(), vx.end());
    for (i = 1; i < vx.size(); i++) {
      if (vx[i - 1].first == vx[i].first)
        cout << "R";
      else
        cout << "D";
    }
    cout << endl;
  }
}
void findxx1(int i, int j) {
  if (i == 0 && j == 0) {
    return;
  } else {
    vx.push_back({par1[i][j].x, par1[i][j].y});
    findxx1(par1[i][j].x, par1[i][j].y);
  }
}
void findxx2(int i, int j) {
  if (i == 0 && j == 0) {
    return;
  } else {
    vx.push_back({par2[i][j].x, par2[i][j].y});
    findxx2(par2[i][j].x, par2[i][j].y);
  }
}

```