#include <cstdio>
const int N = 1e5 + 5, M = 4e5 + 5;
struct E {
    int v, next;
} e[M]; 
int t, n, m, cnt, len, u, v, h[N], ind[N], outd[N], path[M];
bool vis[M]; //代表某条边是否被使用过 
void add(int u, int v) {
    e[++len].v = v; e[len].next = h[u]; h[u] = len;
} 
bool ok() {
    if (t == 1) {
        for (int i = 1; i <= n; i++) {
            if ((ind[i] + outd[i]) & 1) return false;
        }
    } else {
        for (int i = 1; i <= n; i++) {
            if (ind[i] != outd[i]) return false;
        }
    }
    return true;
}
void dfs(int u) {
    //变成引用 等于删除这条边 用完就删 
    for (int &j = h[u]; j; j = e[j].next) {
        if (vis[j]) continue;
        vis[j] = true;
        if (t == 1) vis[j ^ 1] = true; //无向图的反向边 
        //由于要删除边 所以记录编号
        int no = t == 1 ? j / 2 : j - 1; //如果是无向边的话 边数为/ 2 
        if (t == 1 && (j & 1)) no = -no;
        int v = e[j].v;
        dfs(v);
        path[++cnt] = no;
    }
}
int main() {
    scanf("%d%d%d", &t, &n, &m); len = 1; //节点编号从2开始 
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &u, &v);
        add(u, v);
        if (t == 1) add(v, u);
        ind[v]++, outd[u]++;
    }
    //判断是否有解
    if (!ok()) printf("NO");
    else {
        //判断是否找到回路  
        for (int i = 1; i <= n; i++) {
            if (h[i]) { //代表不是孤立点 
                dfs(i);
                break;
            }
        }
        if (cnt != m) printf("NO");
        else {
            printf("YES\n");
            for (int i = cnt; i > 0; i--) printf("%d ", path[i]); 
        } 
    } 
    return 0;
}