///按照题目要求，就是找出一条从左上角到右下角中每个数含2 or 5 最少的路
///可以用Dp的思想， 然后把每个节点该走的方向记下来
///再从终点回溯，把路径存入栈，再输出
///数据会有0的情况， 这时候我们应该记录离终点最近的0
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1000 + 7;
LL Map[maxn][maxn];
LL Dp[maxn][maxn][2];
char Step[maxn][maxn][2];
stack<char> Mesure;int Factor(int Num, int Base) ///得到因子 2 和 5  的个数
{if(Num == 0) return 1;int ret = 0;while(Num % Base == 0){ret++;Num /= Base;}return ret;
}int main()
{ios::sync_with_stdio(false);cin.tie(0);int n;cin >> n;bool Zero = false;int  Zero_Pos;for(int i = 0; i < n; ++i){for(int j = 0; j < n; ++j){cin >> Map[i][j];if(Map[i][j] == 0){Zero = true;Zero_Pos = i;}Dp[i][j][0] = Factor(Map[i][j],2);Dp[i][j][1] = Factor(Map[i][j],5);}}for(int i = 0; i < n; ++i){for(int j = 0; j < n; ++j){for(int k = 0; k < 2; ++k){if(i == 0 && j == 0) continue;if(i == 0){Dp[i][j][k] += Dp[i][j-1][k];Step[i][j][k] = 'R';}else if(j == 0){Dp[i][j][k] += Dp[i-1][j][k];Step[i][j][k] = 'D';}else{Dp[i][j][k] += min(Dp[i-1][j][k],Dp[i][j-1][k]);Step[i][j][k] = Dp[i-1][j][k] < Dp[i][j-1][k] ? 'D' : 'R';}}}}if(min(Dp[n-1][n-1][0],Dp[n-1][n-1][1]) > 1 && Zero){printf("1\n");for(int i = 0; i < Zero_Pos; ++i) putchar('D');for(int i = 0; i < n-1; ++i) putchar('R');for(int i = Zero_Pos; i < n-1; ++i) putchar('D');//
    }else{printf("%d\n",min(Dp[n-1][n-1][0],Dp[n-1][n-1][1]));int k = 0;k = Dp[n-1][n-1][0] < Dp[n-1][n-1][1] ? 0 : 1;for(int i = n-1, j = n-1; i != 0|| j != 0; ){Mesure.push(Step[i][j][k]);if(Step[i][j][k] == 'D') i--;else j--;}while( !Mesure.empty() ) putchar(Mesure.top()), Mesure.pop();}puts("");return 0;
}