#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<iostream>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define O_(x) cout << #x << " " << x << "  ";
#define B cout << "breakpoint" << endl;
typedef long long ll;
typedef pair<ll,int> pii;
const ll inf = 1e14;
#define mp make_pair
inline int read()
{int ans = 0,op = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-') op = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){(ans *= 10) += ch - '0';ch  = getchar();}return ans * op;
}
const int maxn = 303;
const int N = 1e6 + 5;
int n,m;
int id[maxn][maxn];//id[i][j] : i行j列对应线段树中哪个节点
int ls[N],rs[N],root[maxn],tot;
struct egde
{int to,next;ll cost;
}e[N * 20];
int fir[N],alloc;
void adde(int u,int v,int w)
{e[++alloc].next = fir[u];fir[u] = alloc;e[alloc].to = v;e[alloc].cost = w;
}
int build(int line,int i,int l,int r)
{if(l == r){id[line][l] = i;return i;}int mid = l + r >> 1;ls[i] = build(line,++tot,l,mid);rs[i] = build(line,++tot,mid + 1,r);adde(i,ls[i],0); adde(i,rs[i],0);return i;
}
void add(int i,int l,int r,int u,int ql,int qr,ll w)//u->[ql,qr]
{if(l == ql && r == qr){adde(u,i,w);return;}int mid = l + r >> 1;if(qr <= mid) add(ls[i],l,mid,u,ql,qr,w);else if(ql > mid) add(rs[i],mid + 1,r,u,ql,qr,w);else add(ls[i],l,mid,u,ql,mid,w),add(rs[i],mid + 1,r,u,mid + 1,qr,w);
}
bool vis[N];
ll dis[N];
void dij(int s)
{for(int i = 1;i <= tot;i++) dis[i] = inf;memset(vis,0,sizeof(vis));dis[s] = 0;priority_queue<pii,vector<pii>,greater<pii> > p;p.push(mp(dis[s],s));while(p.size()){pii t = p.top();p.pop();int u = t.second;if(vis[u]) continue;vis[u] = 1;for(int i = fir[u];i;i = e[i].next){int v = e[i].to;ll w = e[i].cost;if(dis[v] > dis[u] + w){dis[v] = dis[u] + w;if(!vis[v]) p.push(mp(dis[v],v));}}}
}
int b[maxn][maxn];
ll a[maxn][maxn];
int main()
{n = read(),m = read();for(int i = 1;i <= n;i++) root[i] = build(i,++tot,1,m);for(int i = 1;i <= n;i++)for(int j = 1;j <= m;j++)b[i][j] = read();//距离 for(int i = 1;i <= n;i++)for(int j = 1;j <= m;j++)scanf("%lld",&a[i][j]);//费用for(int i = 1;i <= n;i++)for(int j = 1;j <= m;j++){for(int k = max(1,i - b[i][j]);k <= min(n,i + b[i][j]);k++){int res = b[i][j] - abs(k - i);add(root[k],1,m,id[i][j],max(1,j - res),min(m,j + res),a[i][j]);}}int x1 = read(),y1 = read(),x2 = read(),y2 = read(),x3 = read(),y3 = read();int x = id[x1][y1],y = id[x2][y2],z = id[x3][y3];dij(x);ll xy = dis[y],xz = dis[z];dij(y);ll yz = dis[z],yx = dis[x];dij(z);ll zx = dis[x],zy = dis[y];ll ansx = yx + zx,ansy = xy + zy,ansz = xz + yz;if(ansx >= inf && ansy >= inf && ansz >= inf){printf("NO\n");return 0;}if(ansx <= ansy && ansx <= ansz) printf("X\n%lld",ansx);else if(ansy <= ansx && ansy <= ansz) printf("Y\n%lld",ansy);else printf("Z\n%lld",ansz);
}