#include <bits/stdc++.h>
using namespace std;
#define fo(v, a, b) for(int v = a; v <= b; v++)
#define fr(v, a, b) for(int v = a; v >= b; v--)
#define cl(a, v) memset(a, v, sizeof(a))

const int N = 82, M = 6410;

int n, m, a[N], b[N];
int dp[N][N][N];

void solve(int l, int r, int x) {
    if(l == r) return ;
    int L = r - l, now = 0;
    while(true) {
        fo(p, l, r - 1) {
            int A = p - l, B = r - 1 - p;
            if(((A == 0 && x == b[l]) || (A && dp[l][p][x % A] >= x)) &&
                ((B == 0 && x == b[r]) || (B && dp[p + 1][r][x % B] >= x))) {
                solve(l, p, x), solve(p + 1, r, x);
                fo(i, l, r - 1) a[i] += now;
                return ;
            }
        }
        x += L, now++;
    }
}

inline int lcm(int x, int y) {
    if(!x || !y) return x | y;
    return x / __gcd(x, y) * y;
}
inline void upd(int &x, int y) {  x = max(x, y);  }
inline int calc(int Z, int i, int lim) {
    int c = (lim / Z) * Z + i;
    return c > lim ? c - Z : c;
}

int val[N];

int main()
{
    scanf("%d", &n); fo(i, 1, n) scanf("%d", &b[i]);
    if(n == 1) {
        if(b[1]) puts("No"); else puts("Yes"), puts("");
        exit(0);
    }

    cl(dp, 0xff);
    fo(len, 2, n) fo(l, 1, n - len + 1) {
        int r = l + len - 1, L = r - l;
        if(len == 2) {
            if(b[l] == b[r]) dp[l][r][0] = b[l];
            continue;
        }
        fo(p, l, r - 1) {
            int A = p - l, B = r - 1 - p;
            if(A == 0) {
                int v = b[l];
                if(dp[p + 1][r][v % B] >= v) upd(dp[l][r][v % L], v);
                continue;
            }
            if(B == 0) {
                int v = b[r];
                if(dp[l][p][v % A] >= v) upd(dp[l][r][v % L], v);
                continue;
            }

            int C = lcm(A, B); cl(val, 0xff);
            for(int i = 0; i < C; i++) {
                int cur = min(calc(C, i, dp[l][p][i % A]), calc(C, i, dp[p + 1][r][i % B]));
                if(cur >= 0) upd(val[cur % L], cur);
            }
            fo(t, 0, 1) fo(i, 0, L - 1) upd(val[(i - C) % L], val[i] - C);
            fo(i, 0, L - 1) upd(dp[l][r][i], val[i]);
        }
    }

    if(dp[1][n][0] < 0) puts("No"), exit(0);
    puts("Yes"), solve(1, n, 0);
    fo(i, 1, n - 1) printf("%d ", a[i]);
    puts("");

    return 0;
}