|F_k|=\sum_{E\in F_k}(w^{-}(E)+1)\leq \sum_{E\in
F_k}w^{+}(E)=w^{+}(F_k),\ \ \ \ k\geq 2.$$ Hence, $$w^{-}(F_1)=w^{+}(F_2)=\sum_{k\geq2}(w^{+}(F_k)-w^{-}(F_k))\geq
\sum_{k\geq2}|F_k|=|F/F_1|.$$ Let $g=\max_{E\in F_1} w^{-}(E)$. Since $w^{-}(F_1)=\sum_{E\in F_1}w^{-}(E)\leq
 g|F_1|$, then $$|F|/|F_1|\leq g+1.$$ The weight function $w: K \rightarrow \mathcal {R}^{+}$ defined as follows. For any arc $(E,E^{'})$ with $E^{'}\in F_k $, we know $E^{'}$ is obtained by coalescing two cycles of $E$, and suppose the length of these two cycles are $l_{1}$ and $l_2$, then define $w(E,E^{'})=(l^{-1}_1+l^{-1}_2)f(k)$, where $f(k)$ is defined as in Lemma 8. Then we have the following two claims.\
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**Claim 1**. *For any $E\in F_k$, $k\geq 2$, $w^{+}(E)\geq
(2\alpha-1.5)n\beta f(k)\log n+2$.*\
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**Claim 2**. *For any $E\in F_k$, $k\geq 1$, $w^{-}(E)\leq n\log
nf(k)$*.\
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By these two claims, set $\beta=1/(2\alpha-1.5)$. Then for $E\in
F_k$, $k\geq 2$, we have $w^{+}(E)-w^{-}(E)\geq 2\geq 1$ and $g=\max_{E\in F_1} w^{-}(E)\leq n\log nf(1)\leq
(2\alpha-1.5)n^{1+1/(2\alpha-1.5)}$. Hence $|F|/|F_1|\leq g+1=
O(n^{1+1/(2\alpha-1.5)})$, which completes the proof.  $\Box$\
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**Proof of Claim 1.** Let $E\in F_k$ be an 1-factor with $k$ cycles $\gamma_1, \cdots, \gamma_k$, of lengths $n_1, n_2, \cdots,
n_k$, $k\geq 2$. We proceed to bound $w^{+}(E)$. To show the lower bound of $w^{+}(E)$, we need to count the number of arcs leaving $E$. Suppose $(E,E^{'})$ to be such an arc. Let $\gamma=E\oplus
E^{'}$, $E^{'}\in 